What are the possible values of the sum of squares, given a specific sum?

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In summary, the sum of squares is a measure of variability or spread in a data set, calculated using the formula Σ(xi - x̄)². It cannot be negative and is directly related to variance and standard deviation. However, it can be heavily influenced by extreme values and may not be suitable for non-normally distributed data. The sum of squares is commonly used in statistical analysis, including ANOVA and regression, to compare data sets and identify outliers.
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Ackbach
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Here is this week's POTW:

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Let $A$ be a positive real number. What are the possible values of $\displaystyle\sum_{j=0}^\infty x_j^2$, given that $x_0,x_1,\ldots$ are positive numbers for which $\displaystyle\sum_{j=0}^\infty x_j=A$?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Re: Problem Of The Week # 259 - Apr 17, 2017

This was Problem A-1 in the 2000 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

The possible values comprise the interval $(0, A^2)$.

To see that the values must lie in this interval, note that
\[
\left(\sum_{j=0}^m x_j\right)^2
= \sum_{j=0}^m x_j^2 + \sum_{0\leq j<k\leq m} 2x_jx_k,
\]
so $\sum_{j=0}^m x_j^2 \leq A^2 - 2x_0x_1$. Letting $m \to \infty$, we have $\sum_{j=0}^\infty x_j^2 \leq A^2-2x_0x_1 < A^2$.

To show that all values in $(0, A^2)$ can be obtained, we use geometric progressions with $x_1/x_0 = x_2/x_1 = \cdots = d$ for variable $d$. Then $\sum_{j=0}^\infty x_j = x_0/(1-d)$ and
\[
\sum_{j=0}^\infty x_j^2 = \frac{x_0^2}{1-d^2} = \frac{1-d}{1+d} \left(
\sum_{j=0}^\infty x_j \right)^2.
\]
As $d$ increases from 0 to 1, $(1-d)/(1+d)$ decreases from 1 to 0. Thus if we take geometric progressions with $\sum_{j=0}^\infty x_j = A$, $\sum_{j=0}^\infty x_j^2$ ranges from 0 to $A^2$. Thus the possible values are indeed those in the interval $(0, A^2)$, as claimed.
 

FAQ: What are the possible values of the sum of squares, given a specific sum?

What is the formula for calculating the sum of squares?

The formula for calculating the sum of squares is Σ(xi - x̄)², where xi represents each individual value in a data set and x̄ represents the mean or average of the data set.

Can the sum of squares be negative?

No, the sum of squares cannot be negative. It is a measure of the variability or spread of a data set and is always a non-negative value.

How does the sum of squares relate to variance and standard deviation?

The sum of squares is directly related to both variance and standard deviation. In fact, variance is the average of the sum of squares, and the standard deviation is the square root of the variance.

Are there any limitations to using the sum of squares as a measure of variability?

The sum of squares can be heavily influenced by extreme values in a data set, which can skew its value. Additionally, it may not be an appropriate measure of variability for non-normally distributed data.

How can the sum of squares be used in statistical analysis?

The sum of squares is an important component in many statistical tests, such as ANOVA and regression analysis. It can also be used to compare the variability of different data sets or to identify outliers in a data set.

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